Prove that $x=0$ is the unique solution to the following equation

Question

The equation is given as follows: $e^x - (1+x + \frac{x^2}{2!} + \frac{x^3}{3!}+...+\frac{x^n}{n!}) =0 $

In order to prove the uniqueness I tried using the Integral form of the remainder (since the equation can be represented in the form $f(x) - P_n(x) = R_n(x))$, but I'm not sure whether this is the right way to proceed.

Answer 1

We have $R_n(x)=\dfrac{e^{c_x}}{(n+1)!}x^{n+1}$ for some $c_x\in\bigl[0,|x|\bigr].$ It is easy to see that $R_n(x)=0\iff x=0.$

Answer 2

With the integral remainder, $$e^x-\sum_{k=0}^n \frac{x^k}{k!}=\frac 1{n!}\int_0^x (x-t)^n e^t dt$$

$t\mapsto (x-t)^n e^t$ does not change sign over $[\min(0,x),\max(0,x)]$, hence $e^x-\sum_{k=0}^n \frac{x^k}{k!}=0$ implies $\forall t\in [\min(0,x),\max(0,x)], (x-t)^n=0$ which can only happen when $x=0$.

Answer 3

By contradiction assume there is another zero $c_0$ of $$f_n(x)=e^x - (1+x + \frac{x^2}{2!} + \frac{x^3}{3!}+...+\frac{x^n}{n!})$$

so by Rolle's theorem $f'_n=f_{n-1}$ has a zero $c_{1}$ between 0 and $c_0$ and by induction $f^{(n+1)}=\exp$ has also a zero $c_{n+1}$ between 0 and $c_n$ which is a contradiction.